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%%文档的题目、作者与日期
%\author{王立庆（2024级数学与应用数学1班）}
\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{高等代数测验1 - 行列式 - 解答}
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\date{2024 年 10 月 24 日}

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\begin{document}

\maketitle


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\begin{enumerate}\itemsep0em

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\item %1
按定义计算 5 阶行列式 
$D=\begin{vmatrix}
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 3 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 4 \\
5& 0 & 0 & 0 & 0
\end{vmatrix}$ 的值。

\vspace{0.2cm}

{\color{red} 解答：只有 $a_{14}a_{23}a_{32}a_{45}a_{51}$ 这一项不为零，因此计算列指标的逆序数，可得
$$D=(-1)^{\tau(43251)}(1)(2)(3)(4)(5)=(-1)^7120=-120.$$

}


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\item %2 
求行列式  
$f(x)=\begin{vmatrix}
2 & 0 & x & 4 \\
1 & 0 & 2 & 4 \\
1 & 2 & 5 & 5 \\
1 & 3 & 4 & 0
\end{vmatrix}$ 的 $x$ 的系数。 

\vspace{0.2cm}

{\color{red} 解答：这个行列式只在 (1,3) 位置有 $x$, 因此按第一行展开，可得 $x$ 的系数为
\[ 
(-1)^{1+3}\begin{vmatrix}
1 & 0 & 4 \\
1 & 2 & 5 \\
1 & 3 & 0
\end{vmatrix} = -11. 
\]

%\begin{lstlisting}[language=R]
%> A=matrix(c(1,0,4,1,2,5,1,3,0),nrow=3,byrow=T)
%> print(det(A))
%\end{lstlisting}
}

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\item  %3
计算下述行列式的值：
\begin{equation*}
D=\begin{vmatrix}
1+a & 1 & 1 & 1  \\ 
1 & 1+b & 1 & 1  \\ 
1 & 1 & 1+c & 1  \\ 
1 & 1 & 1 & 1+d  \\ 
\end{vmatrix}
\end{equation*}

\vspace{0.2cm}

{\color{red} 解答一：在上边和左边增加一行一列，然后用第一行乘以 $-1$ 分别加到第2-5行，可得

\begin{equation*}
D=\begin{vmatrix}
1&1&1&1&1 \\ 
0&1+a & 1 & 1 & 1  \\ 
0&1 & 1+b & 1 & 1  \\ 
0&1 & 1 & 1+c & 1  \\ 
0&1 & 1 & 1 & 1+d  \\ 
\end{vmatrix}
=\begin{vmatrix}
1&1&1&1&1 \\ 
-1&a & 0 & 0 & 0  \\ 
-1&0 & b & 0 & 0  \\ 
-1&0 & 0 & c & 0  \\ 
-1&0 & 0 & 0 & d  \\ 
\end{vmatrix}.
\end{equation*}
再用第2-5行分别化简第一行，（设 $a,b,c,d\neq 0$），可得
\begin{equation*}
D=
=\begin{vmatrix}
1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}&0&0&0&0 \\ 
-1&a & 0 & 0 & 0  \\ 
-1&0 & b & 0 & 0  \\ 
-1&0 & 0 & c & 0  \\ 
-1&0 & 0 & 0 & d  \\ 
\end{vmatrix}
=\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)abcd.
\end{equation*}

}

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\item  %4
使用初等行变换，将行列式化为上三角行列式，计算行列式的值 
$D=\begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 3 & 4 & 5 \\
1 & 4 & 7 & 13 \\
\end{vmatrix}$ 的值。

{\color{red} 解答：用行初等变换，化为上三角行列式，可得 
\begin{equation*}
D=\begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 3 & 4 & 5 \\
1 & 4 & 7 & 13 \\
\end{vmatrix} 
= 
\begin{vmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 2 & 3 & 4 \\
0 & 3 & 6 & 12 \\
\end{vmatrix} 
= 
\begin{vmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 2 \\
0 & 0 & 3 & 9 \\
\end{vmatrix} 
=
\begin{vmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 3 \\
\end{vmatrix} 
=3. 
\end{equation*}

%\begin{lstlisting}[language=R]
%> A=matrix(c(1,1,1,1,1,2,2,2,1,3,4,5,1,4,7,13),nrow=4,byrow=T)
%> print(det(A))
%\end{lstlisting}

}



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\item %5
考虑下述行列式，记 $A_{ij}$ 为元素 $a_{ij}$ 对应的代数余子式，计算表达式 $A_{11}+A_{21}+A_{31}+A_{41}$ 的值。
\begin{equation*}
D=|a_{ij}| = \begin{vmatrix}
2&3&2&3\\
4&5&4&5\\
7&6&0&0\\
9&8&0&0\\
\end{vmatrix}. 
\end{equation*}

\vspace{0.2cm}

{\color{red} 解答：根据行列式按第一列展开的计算方法，可得
$A_{11}+A_{21}+A_{31}+A_{41}$ 是将行列式 $D$ 的第一列换成1,1,1,1得到的行列式的值。因此所求为
\begin{equation*}
\begin{vmatrix}
1&3&2&3\\
1&5&4&5\\
1&6&0&0\\
1&8&0&0\\
\end{vmatrix}
= 
\begin{vmatrix}
2&3\\
4&5\\
\end{vmatrix}
\begin{vmatrix}
1&6\\
1&8\\
\end{vmatrix}
= (10-12)(8-6)=-4. 
\end{equation*}

}

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\item %6 
写出克拉默法则，求解线性方程组，
\begin{equation*}
\left\{
\begin{aligned}
2x + 3y - z &= 1, \\ 
x - y + 2z &= -2, \\ 
3x - 2y + 4z &= 3. 
\end{aligned} 
\right. 
\end{equation*}

\vspace{0.2cm}

{\color{red} 解答：系数行列式等于
\begin{equation*}
D=\begin{vmatrix}
2&3&-1 \\ 
1&-1&2 \\ 
3&-2&4 \\ 
\end{vmatrix} = 5.
\end{equation*}
分别将系数行列式的第1-3列换成方程右边的常数，得到三个行列式为
\begin{equation*}
D_1=\begin{vmatrix}
1&3&-1 \\ 
-2&-1&2 \\ 
3&-2&4 \\ 
\end{vmatrix} = 35, \,\,\, 
D_2=\begin{vmatrix}
2&1&-1 \\ 
1&-2&2 \\ 
3&3&4 \\ 
\end{vmatrix} = -35, \,\,\, 
D_3=\begin{vmatrix}
2&3&1 \\ 
1&-1&-2 \\ 
3&-2&3 \\ 
\end{vmatrix} = -40.
\end{equation*}
根据克拉默法则，可得这个线性方程组有唯一解
\begin{equation*}
\left\{
\begin{aligned}
x &=\frac{D_1}{D} = \frac{35}{5} = 7,\\ 
y &=\frac{D_2}{D} = \frac{-35}{5} = -7,\\ 
z &=\frac{D_3}{D} = \frac{-40}{5} = -8. \\ 
\end{aligned} 
\right. 
\end{equation*}

}


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\end{enumerate}

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\end{document}
